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#rainbow[ *Solution* (answer $f(0, 1) = 5$): There are a few ways to approach this problem. *Solution 1*: Note that $f(x_1, x_2) = 3x_1^2 + 5x_2^2 = 3(x_1^2 + x_2^2) + 2x_2^2 = 3 + 2x_2^2$. Since $x_2^2 <= 1$ and equality holds when $(x_1, x_2) = (0, 1)$, $f$ is maximized at $f(0, 1) = 5$. *Solution 2*: Let $a_1 = x_1^2$, $a_2 = x_2^2$. Thus, we wish to maximize $ f(x_1, x_2) = 3x_1^2 + 5x_2^2 = 3a_1 + 5a_2 $ subject to $x_1^2 + x_2^2 = 1$ or $a_1 + a_2 = 1$ and $a_1 >= 0$, $a_2 >= 0$. Intuitively, we want to make $a_2$ as large as possible. Note that $3a_1 + 5a_2 >= 5a_1 + 5a_2 = 5$, since $a_1 >= 0$ and equality holds when $(a_1, a_2) = (0, 1)$. Thus, $f$ is maximized at $f(0, 1) = 5$. ]

#rainbow[ *Solution* (answer $f(0, 1) = 5$): Let $vr(x) = vec(x_1, x_2)$. Our objective function rewrites itself to: $ f(x_1, x_2) = vecrow(x_1, x_2) mat(3, 0; 0, 5;) vec(x_1, x_2) = 3x_1^2 + x_2^2 $ Our constraint $norm(vr(x)) = 1$ rewrites itself to $sqrt(x_1^2 + x_2^2) = 1 => x_1^2 + y_1^2 = 1$. Observe that this is exactly the same as in *Problem 1*, so our answer is once again $5$. ]

#rainbow[ *Solution* (answer $f(0, 1, 0) = 4$): Let $vr(x) = vec(x_1, x_2, x_3)$. Our objective function rewrites itself to: $ f(x_1, x_2, x_3) = vecrow(x_1, x_2, x_3) mat(1, 0, 0; 0, 4, 0; 0, 0, 2;) vec(x_1, x_2, x_3) = x_1^2 + 4x_2^2 + 2x_3^2 $ Our constraint is $x_1^2 + x_2^2 + x_3^2 = 1$. Again, it intuitively makes sense to make $x_2$ as large as possible. Indeed, we have $x_1^2 + 4x_2^2 + 2x_3^2 <= 4x_1^2 + 4x_2^2 + 4x_3^2 = 4$, and equality holds at $(0, 1, 0)$. Thus, $f$ is maximized at $f(0, 1, 0) = 4$. ]

#rainbow[ *Theorem 1*: Let $D$ be a diagonal n by n matrix. Say we wish to maximize $f(vr(x)) = vr(x)^TT A vr(x)$ subject to $norm(vr(x)) = 1$. Let $m = max_i D_(i,i)$ be the maximal diagonal entry of $D$, and let $j$ be any integer satisfying $m = D_(j, j)$. The maximal value of $f$ under these constraints is $x$, achieved at $x_1 = x_2 = dots.h = x_(j - 1) = x_(j + 1) = dots.h = x_(n - 1) = x_n = 0$ and $x_j = 1$. ]

#hint[$A$ is symmetric. This is not a coincidence.] #hint[We can write $A = P D P^TT$ for orthogonal matrix $P$ and diagonal matrix $D$.] #hint[Say we wish to maximize $vr(x)^TT P D P^TT vr(x)$. What if we tried letting $u = P^TT vr(x)$. What is $u^TT$?] #hint[Now we want to maximize $vr(u)^TT D vr(u)$. Looks awfully familiar, doesn't it? How can we find $vr(x)$ from $vr(u)$?] #hint[(for the bonus): When we maximize $f$, $vr(u)$ is a very special vector. Is there anything special about $vr(x)$?]

#rainbow[ *Solution* (answer $f(1/sqrt(2), 1/sqrt(2)) = 5$): $A$ is symmetric because $A = A^TT$. Thus, there exists orthogonal matrix $P$ and diagonal matrix $D$ such that $A = P D P^TT$. Calculating, we get $ P = mat(-1/sqrt(2), 1/sqrt(2); 1/sqrt(2), 1/sqrt(2);), D = mat(3, 0; 0, 5) $ $ f(vr(x)) = vr(x)^TT A vr(x) = vr(x)^TT P D P^TT vr(x) $ Now, let $vr(u)$ = $P^TT vr(x)$. Since $vr(u)^TT = (P^TT vr(x))^TT = vr(x)^TT P^TT^TT = vr(x)^TT P$, $ f(vr(x)) = vr(u)^TT D vr(u) = vr(u)^TT mat(3, 0; 0, 5;) vr(u) $ Furthermore, $norm(vr(u)) = norm(P^TT vr(x)) = norm(vr(x)) = 1$ since $P$ is orthogonal. Note that this is exactly *Problem 2*! The function is maximized when $vr(u) = vec(0, 1)$ with $f(vr(u)) = 5$. Now to obtain $vr(x)$, $vr(u) = P^TT vr(x)$, so $P vr(u) = P P^TT vr(x) = I_n vr(x) = vr(x)$, since $P$ is orthogonal. Thus, $vr(x) = P vr(u) = vec(1/sqrt(2), 1/sqrt(2))$. ]

There's actually something special about both the maximal value and the corresponding value of $vr(x)$. #rainbow[ *Theorem 2*: Let $A$ be a symmetric matrix, $P$ an orthogonal matrix, and $D$ a diagonal matrix satisfying $A = P D P^TT$. Say we want to maximize $f(vr(x)) = vr(x)^TT A vr(x)$ subject to $norm(vr(x)) = 1$. Let $m$ be the maximal eigenvalue of $A$. The maximal value of $f$ is $m$, and any eigenvector $vr(v)$ corresponding to $m$ satisfies $f(vr(v)) = m$. ] This is true because all diagonal entries of $D$ are eigenvalues of $A$, and by *Theorem 1* the maximal value of $vr(u)^TT D vr(u)$ subject to $norm(u) = 1$ is the maximal diagonal entry of $D$. The special property of $vr(u)$ being all $0$s and a $1$ means that $vr(x) = P vr(u)$ is a column of $P$, which is an eigenvector corresponding to the eigenvalue. Now we'll see how we can reduce the case for general $A$ to the symmetric case.

#rainbow[ *Solution* (answer $4x_1^2 + 2x_1x_2 + 4x_2^2$): $ vr(x)^TT mat(4, 1; 1, 4;) vr(x) &= vecrow(x_1, x_2) mat(4, 1; 1, 4;) vec(x_1, x_2)\ &= vecrow(4x_1 + x_2, x_1 + 4x_2) vec(x_1, x_2)\ &= 4x_1^2 + x_1x_2 + x_1x_2 + 4x_2^2\ &= 4x_1^2 + 2x_1x_2 + 4x_2^2 $ Similarly, $ vr(x)^TT mat(4, 0; 2, 4;) vr(x) = 4x_1^2 + 2x_1x_2 + 4x_2^2 $ Observe that these two are the same. ]

#rainbow[ *Lemma 1*:Let $A$ be a $n$ by $n$ matrix and $vr(x) = vec(x_1, x_2, dots.h, x_n)$. Then, $ vr(x)^TT A vr(x) = sum_(1 <= i, j <= n) A_(i, j) x_i x_j $ ] This is left as an exercise to the reader #emoji.face.tongue. #rainbow[ *Lemma 2*: If $A$, $B$ are $n$ by $n$ matrices satisfying $A + A^TT = B + B^TT$, $vr(x)^TT A x = vr(x)^TT B vr(x)$ for all vectors $vr(x) in RR^n$ ] Consider the coefficient $c_(i, j)$ of $x_i x_j$ from *Lemma 1*. If $i != j$, then $ c_(i, j) = A_(i, j) + A_(j, i) = B_(i, j) + B_(j, i) $ since $A + A^TT = B + B^TT$. Also, $c_(i, i) = A_(i, i) = B_(i, i)$. Thus, the expansions of both $vr(x)^TT A vr(x)$ and $vr(x)^TT B vr(x)$ are the same and the expressions are equal. Let $B = (A + A^TT) / 2$. $ A + A^TT = B + B^TT = (A + A^TT) / 2 + ((A + A^TT) / 2)^TT = A+A^TT $ so optimizing $vr(x)^TT A vr(x)$ is the same as optimizing $vr(x)^TT B vr(x)$ by *Lemma 2*. Since $B$ is symmetric, we've successfully reduced the general case to the symmetric case.

= Connection to Calc 3 Did you see multivariable optimization and think Calc 3 lagrange multipliers? It turns out there actually is a rather interesting connection between lagrange multipliers and eigenvalues. Let's approach the general problem of optimizing $vr(x)^TT A vr(x)$ subject to $norm(vr(x)) = 1$ using Calc 3. Let's assume $A$ is upper triangular, and we can reduce the general version to the upper triangular case. Thus, our objective function becomes $ f(vr(x)) = sum_(1 <= i <= j <= n) A_(i, j) x_i x_j $ Our constraint function is $ g(vr(x)) = x_1^2 + x_2^2 + dots.h + x_n^2 = 1 $ Applying lagrange multipliers, we want $grad f = lambda grad g$. Let $C = A + A^TT$. Thus, $C_(i, j) = C_(j, i) = A_(i, j)$ for $i < j$ and $C_(i, i) = 2A_(i, i)$ $ pdv(f, x_i) &= pdv(, x_i) A_(1, i) x_1 x_i + A_(2, i) x_2 x_i + dots.h + A_(i - 1, i) x_(i - 1) x_i + A_(i, i) x_i^2 + A_(i, i + 1) x_i x_(i + 1) + dots.h + A_(i, n) x_i x_n\ &= A_(1, i) x_1 + A_(2, i) x_2 + dots.h + A_(i - 1, i) x_(i - 1) + 2A_(i, i) x_i + A_(i, i + 1) x_(i + 1) + dots.h + A_(i, n) x_n\ &= C_(i, 1) x_1 + C_(i, 2) x_2 + dots.h + C_(i, n) x_n = C_i vr(x) $ $ grad f = vec(pdv(f, x_1), pdv(f, x_2), dots.v, pdv(f, x_n)) = vec(C_1 vr(x), C_2 vr(x), dots.v, C_n vr(x)) = C vr(x) $ $ pdv(g, x_i) = 2 x_i $ $ grad g = vec(2x_1, 2x_2, dots.v, 2x_n) = 2vr(x) $ Since we want $grad f = lambda grad g$, $ grad f &= lambda grad g\ C vr(x) &= lambda 2vr(x)\ C/2 vr(x) &= lambda vr(x) $ Note that this is exactly the condition for $vr(x)$ being an eigenvector of $C/2$. Thus, the function is maximized at an eigenvector of $(A + A^TT) / 2$. \ \